suppose that $G$,$H$,$K$ are connected graphs with at least two vertices,prove that the only solution of the equation $L(G)=H \square K$ is $K=K_n$ ,$H=K_m$ and $G=K_{m,n}$.
because the eigenvalue of any line graph $L(G)$ is not fewer than -2 (as a theorem),if we suppose that $K$ and $H$ will be some thing else,some how I must make a contradiction. but I can't make it.
I don't what to do,any hint or guidance me to be in right way will be great ,thanks.
You do not make clear if you are required to use an algebraic approach. There is a simple `elementary' solution.
Assume vertices $x$ and $y$ in $H$ are not adjacent. Let $P$ be a shortest $x,y$-path in $H$, $P=(x,a,b,\ldots,y)$ (where $a\ne y$, but $b$ may be equal to $y$). Now let $p$ and $q$ be two adjacent vertices in $K$ (these exist, because $K$ is connected and has at least two vertices).
Then the four vertices $(x,p),(a,p),(b,p),(a,q)$ of $H\square K$ induce a claw ($K_{1,3}$) and it is a trivial exercise to show that a linegraph cannot contain an induced claw.
So all vertices of $H$ must be adjacent. Now you can finish the proof :)